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\mathrm{\\{} E X P L ~ B e s i d e s ~ p r o v i d i n g ~ a n ~ e a s y ~ w a y ~ t o ~ d i f f e r e n t i a t e ~ p r o d u c t s , ~ l o g . ~ arithmic differentiation also provides a measure of the relative or fractional rate of change, defined as $y^{\prime} / y . Show that the relative rate of change of any polynomial approaches zero as the independent variable approaches infinity.

Short Answer

Expert verified
The relative rate of change of any polynomial approaches zero as \( x \to \infty \).

Step by step solution

01

Understand the Concept of Relative Rate of Change

The relative rate of change of a function \( y \) is defined as \( \frac{y'}{y} \). This measures how fast a function changes relative to its current value.
02

Consider a General Polynomial Function

Let's take a general polynomial function of degree \( n \), given by \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \), where \( a_n eq 0 \).
03

Differentiate the Polynomial

The derivative of \( f(x) \) is given by \( f'(x) = na_nx^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1 \).
04

Set Up the Relative Rate of Change Expression

To find the relative rate of change, we calculate \( \frac{f'(x)}{f(x)} = \frac{na_nx^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1}{a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0} \).
05

Simplify for Large Values of \( x \)

For large \( x \), the highest degree term \( a_nx^n \) dominates both the numerator and denominator. Thus, simplifying gives \( \frac{f'(x)}{f(x)} \approx \frac{na_nx^{n-1}}{a_nx^n} = \frac{n}{x} \).
06

Evaluate the Limit as \( x \to \infty \)

As \( x \to \infty \), \( \frac{n}{x} \to 0 \). Thus, the relative rate of change \( \frac{f'(x)}{f(x)} \) approaches zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Function
A polynomial function is one of the fundamental building blocks in algebra and calculus. It is expressed in the form \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \). Here, \( a_n, a_{n-1}, \ldots, a_0 \) are constant coefficients, and \( n \) is a non-negative integer that represents the degree of the polynomial. - The degree of the polynomial is critical because it tells us about the behavior of the function as \( x \) gets very large or very small.- The highest power of \( x \), which is \( x^n \) in this case, dominates as \( x \to \infty \).These functions can graphically represent curves and are widely used to model real-world phenomena because of their simple yet flexible nature. As \( x \) approaches infinity, higher-degree terms overpower the smaller ones, leading us to make generalizations about the function's behavior.
Logarithmic Differentiation
Logarithmic differentiation is a powerful technique in calculus used primarily when dealing with complex multiplicative functions. It simplifies the differentiation process by taking the natural logarithm of a function, which turns products into sums and rational expressions into differences.- For a function \( y = f(x) \), we take the natural logarithm of both sides: \( \ln(y) = \ln(f(x)) \).- We then differentiate both sides using implicit differentiation. This technique is very useful for finding relative rates of change.After differentiating, solving for \( y' \) gives us the derivative of the original function. This approach is particularly beneficial when dealing with products of functions or functions raised to powers, where direct differentiation might be daunting.
Calculus
Calculus is the branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It serves as the foundation for most advanced mathematics and a multitude of scientific discoveries. - **Differential Calculus** is concerned primarily with the concept of a derivative, which measures how a function changes as its input changes. - **Integral Calculus** focuses on accumulation of quantities and the areas under and between curves. Within calculus, we use tools like derivatives to analyze the behavior of functions, determine their rates of change, and solve complex problems. Understanding calculus is essential, as it plays a critical role in modeling and solving real-world problems in physics, engineering, economics, and beyond.
Differentiation Techniques
Differentiation techniques are various methods used to find the derivative of functions efficiently. Each technique is suited to different types of functions and scenarios. Here are some common techniques:
  • **Power Rule:** Used when differentiating functions of the form \( x^n \). The derivative is \( nx^{n-1} \).
  • **Product Rule:** Useful when differentiating a product of two functions \( u(x) \) and \( v(x) \). The derivative is \( u'v + uv' \).
  • **Quotient Rule:** Applicable in differentiating a ratio of two functions. Given \( \frac{u(x)}{v(x)} \), the derivative is \( \frac{u'v - uv'}{v^2} \).
  • **Chain Rule:** Used when differentiating a composite function, \( f(g(x)) \). The derivative is \( f'(g(x))g'(x) \).
These techniques help simplify differentiation across a broad range of functions, making them indispensable in calculus.

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Most popular questions from this chapter

The arithmetic mean of the numbers \(a\) and \(b\) is \((a+b) / 2\), and the geometric mean of two positive numbers \(a\) and \(b\) is \(\sqrt{a b} .\) Suppose that \(a>0\) and \(b>0\). (a) Show that \(\sqrt{a b} \leq(a+b) / 2\) holds by squaring both sides and simplifying. (b) Use calculus to show that \(\sqrt{a b} \leq(a+b) / 2 .\) Hint: Consider \(a\) to be fixed. Square both sides of the inequality and divide through by \(b .\) Define the function \(F(b)=(a+b)^{2} / 4 b\). Show that \(F\) has its minimum at \(a\). (c) The geometric mean of three positive numbers \(a, b\), and \(c\) is \((a b c)^{1 / 3} .\) Show that the analogous inequality holds: $$ (a b c)^{1 / 3} \leq \frac{a+b+c}{3} $$ Hint: Consider \(a\) and \(c\) to be fixed and define \(F(b)=\) \((a+b+c)^{3} / 27 b .\) Show that \(F\) has a minimum at \(b=\) \((a+c) / 2\) and that this minimum is \([(a+c) / 2]^{2}\). Then use the result from (b).

The total cost of producing and selling \(x\) units of Xbars per month is \(C(x)=100+3.002 x-0.0001 x^{2} .\) If the production level is 1600 units per month, find the average cost, \(C(x) / x\), of each unit and the marginal cost.

Evaluate \(\int|x| d x\).

Evaluate \(\int \sin ^{2} x d x\).

The ZEE Company makes zingos, which it markets at a price of \(p(x)=10-0.001 x\) dollars, where \(x\) is the number produced each month. Its total monthly cost is \(C(x)=200+4 x-0.01 x^{2} .\) At peak production, it can make 300 units. What is its maximum monthly profit and what level of production gives this profit?

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